1. ## ln and tanh

Evaluate
$\displaystyle \displaystyle \int \frac{dy}{1 - y^2}$

There are essentially two ways to approach this. The first is
$\displaystyle \displaystyle \int \frac{dy}{1 - y^2} = tanh^{-1}(y) + C$

The other is to use partial fractions:
$\displaystyle \displaystyle \int \frac{dy}{1 - y^2} = \frac{1}{2} ln \left | \frac{1 + y}{1 - y} \right | + C$

Now, tanh^(-1) and ln have an overlapping domain and in this region everything is just fine and C = 0. The problem is that the ln solution has a much larger domain than the tanh one.

How can two solutions of an integral have different domains?

-Dan

2. Originally Posted by topsquark
Evaluate
$\displaystyle \displaystyle \int \frac{dy}{1 - y^2}$

There are essentially two ways to approach this. The first is
$\displaystyle \displaystyle \int \frac{dy}{1 - y^2} = tanh^{-1}(y) + C$

The other is to use partial fractions:
$\displaystyle \displaystyle \int \frac{dy}{1 - y^2} = \frac{1}{2} ln \left | \frac{1 + y}{1 - y} \right | + C$

Now, tanh^(-1) and ln have an overlapping domain and in this region everything is just fine and C = 0. The problem is that the ln solution has a much larger domain than the tanh one.

How can two solutions of an integral have different domains?

-Dan
Because is...

$\displaystyle \displaystyle \frac{1}{1-y^{2}}= \frac{1}{2}\ (\frac{1}{1+y}+ \frac{1}{1-y})$ (1)

...You can in any case write...

$\displaystyle \displaystyle \int \frac{dy}{1-y^{2}} = \frac{1}{2}\ (\ln |1+y| - \ln |1-y|)+ c$ (2)

... and the expression (2) is valid both for $\displaystyle |y|<1$ and $\displaystyle |y|>1$. The alternative...

$\displaystyle \displaystyle \int \frac{dy}{1-y^{2}} = \tanh^{-1} y + c$ (3)

... is valid only for $\displaystyle |y|<1$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Originally Posted by chisigma
Because is...

$\displaystyle \displaystyle \frac{1}{1-y^{2}}= \frac{1}{2}\ (\frac{1}{1+y}+ \frac{1}{1-y})$ (1)

...You can in any case write...

$\displaystyle \displaystyle \int \frac{dy}{1-y^{2}} = \frac{1}{2}\ (\ln |1+y| - \ln |1-y|)+ c$ (2)

... and the expression (2) is valid both for $\displaystyle |y|<1$ and $\displaystyle |y|>1$. The alternative...

$\displaystyle \displaystyle \int \frac{dy}{1-y^{2}} = \tanh^{-1} y + c$ (3)

... is valid only for $\displaystyle |y|<1$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
I realize that much. My question is that I had thought the integral of an expression was unique up to an arbitrary constant. Because the domains are different, the tanh^(-1) and ln are different functions. I know of no reason why the tanh^(-1) solution does not also represent the most general solution.

-Dan

4. I don't know why you would say "tanh^{-1}(x)[/tex] is the most general solution". As chisquared said, it is only valid for -1< x< 1. The ln solution is valid for |x|> 1. There are each the correct integral on that region.